Usually one has the matrix and wishes to estimate the eigenvalues, but here it's the other way around: I have the positive eigenvalues of an unknown real positive definite matrix and I would like to say something about it's diagonal elements.
The only result I was able to find is that the sum of the eigenvalues coincides with the trace of the matrix, does anyone know of anything more specific? Or perhaps can point me to any literature that discusses this problem?
Answer
There's probably not a lot more to say in terms of bounds. Look at
\begin{bmatrix}
p & q \\ r & 3-p
\end{bmatrix}
Its characteristic polynomial is
$$
x^2 - 3x + p(3-p) - rq = 0
$$
whose values at $x = 1$ and $x = 2$ are identical. By making $rq = p(3-p) + 2$, for instance, by choosing $r = p(3-p)$ and $q = 2$, we get the value 0 at $x = 1, 2$. That means that $1$ and $2$ are the eigenvalues of the matrix, but as $p$ varies over $\mathbb R$, you get every possible diagonal pair that sums up to $1+2$. In other words: the trace constraint, and no other, holds. Since you can take a whole bunch of $2 \times 2$ blocks like this down the diagonal of a $2n \times 2n$ matrix, there are similar freedoms in larger matrices, although this doesn't prove that the diagonal entries are completely free in the sense above -- that might take another page of work; it merely suggests that this more general statement might be true.
No comments:
Post a Comment